A bear spies some honey and takes off from rest, accelerating at a rate of 2.3 m/s2.v If the honey is 18 m away, how fast will his snout be going when it reaches the treat?
We can use the following equation to solve the problem:
v^2 = u^2 + 2*a*s
where v is the final velocity, u is the initial velocity (which is 0), a is the acceleration, and s is the distance traveled.
v^2 = 0^2 + 2 * 2.3 * 18
v^2 = 0 + 2 * 2.3 * 18
v^2 = 82.8
To find the final velocity, we need to take the square root:
v = sqrt(82.8)
v = 9.1 m/s
So the bear's snout will be going 9.1 m/s when it reaches the honey.
To solve this problem, we can use the equations of motion. The bear starts from rest, so its initial velocity (u) is 0 m/s.
We are given the acceleration (a) of the bear, which is 2.3 m/s^2, and the distance (s) to the honey, which is 18 m. We need to find the final velocity (v) of the bear.
The equation that relates initial velocity (u), final velocity (v), acceleration (a), and distance (s) is:
v^2 = u^2 + 2as
Since the initial velocity (u) is 0, we can simplify the equation to:
v^2 = 2as
Plugging in the values:
v^2 = 2 * 2.3 m/s^2 * 18 m
v^2 = 82.8 m^2/s^2
Now, we take the square root of both sides to find the final velocity (v):
v = √(82.8 m^2/s^2)
v ≈ 9.11 m/s
Therefore, when the bear reaches the honey, its snout will be going at a speed of approximately 9.11 m/s.
To find the final velocity of the bear when it reaches the honey, we can use the equation of motion:
vf^2 = vi^2 + 2ad
Where:
vf = final velocity
vi = initial velocity (in this case, 0 m/s as the bear is at rest)
a = acceleration (2.3 m/s^2)
d = distance (18 m)
Substituting the values into the equation:
vf^2 = 0^2 + 2 * 2.3 * 18
vf^2 = 0 + 41.4 * 18
vf^2 = 745.2
Taking the square root of both sides, we get:
vf = √745.2
vf ≈ 27.28 m/s
Therefore, his snout will be going approximately 27.28 m/s when the bear reaches the honey.